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STRUCTURE OF CADMIUM ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( September 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks discovered by Gell-Mann and Zweig I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838,68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Naturally occurring cadmium (Cd) is composed of 8 isotopes. For two of them, natural radioactivity was observed, and three others are predicted to be radioactive but their decays were never observed, due to extremely long half-life times. The 5 stable isotopes have atomic mass 106, 108, 110, 111, 112, and 114.The two natural radioactive isotopes are Cd-113 (beta decay, half-life is 7.7 × 1015 years) and Cd-116 (two-neutrino double beta decay, half-life is 2.9 × 1019 years). The other three are Cd-106, Cd-108 (double electron capture), and Cd-114 (double beta decay); only lower limits on their half-life times have been set. At least three isotopes – Cd-110, Cd-111, and Cd-112 - are absolutely stable (except, theoretically, to spontaneous fission). Among the isotopes absent in the natural cadmium, the most long-lived are Cd-109 with a half-life of 462.6 days, and Cd-115 with a half-life of 53.46 hours. All of the remaining radioactive isotopes have half-lives that are less than 2.5 hours and the majority of these have half-lives that are less than 5 minutes. STRUCTURE OF Cd-96, Cd-98, Cd-100, Cd-102, Cd-104, Cd-106, Cd-108, Cd-110, Cd-112, Cd-114 AND Cd-116 WITH S = 0 For understanding the structure of this group (which includes the 4 stable isotopes existing from Cd-106 to Cd-114) you must read my STRUCTURE OF Cd-106 . In the following diagram of Cd -96 with S =0 you see that the additional p48 and n48 make the symmetrical alpha particle which contributes to the high symmetry. DIAGRAM OF Cd-96 WITH S = 0 In this structure you see the six horizontal planes of opposite spins like the +HP1, -HP2, +HP3, -HP4, +HP5, and -HP6 along with the horizontal squares of opposite spins like the -HSQ and +HSQ. Here the p45, n45, p47, n47, p46, n46, p48, and n48 make the symmetrical alpha particles of high symmetry. Note that the p41, n41, p42, n42, p43, n43 p44, and n44 of opposite spins are not shown. Also 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' ' ' n40.......p40' ' +HSQ p38..........n38 ' ' n31………p12........n12......p32' ' -HP6 p31....n11.........p11…… n32 ' ' p29....... n10.........p10……n30' ' +HP5 n29……p9..........n9 …….p30 ' ' p47......n27.........p8..........n8.......p28....n48' ' -HP4 n45.....p27......n7..........p7.......n28.........p46 ' ' n47......p25.........n6.........p6.......n26....p48' ' +HP3 p45.....n25……p5........n5……..p26.........n46 ' ' n23………p4........n4……..p24' ' -HP2 p23…….n3…….p3……….n24 ' ' p21.........n2………p2........n22' ' +HP1 n21......p1........n1........p22 ' ' p37......n37 ' ' -HSQ n39......p39 ' Then in the presence of an even number of extra neutrons with opposite spins which fill the blank positions we get the unstable structures of the Cd-98, Cd-100, Cd-102 and Cd-104 because the small number of extra neutrons cannot give enough binding energies to pn bonds for overcoming the pp and nn repulsions. However in the stable structures of Cd-106, Cd-108, Cd-110, Cd-112, and Cd-114 the greater number of extra neutrons gives enough binding energies to pn bonds for overcoming the repulsions. Whereas the two more extra neutrons of Cd-116 (in the absence of blank positions) make single bonds leading to the decay. ' ' STRUCTURE OF Cd-118, Cd-120, Cd-122, Cd-124, Cd-126, Cd-128, Cd-130, AND Cd-132 WITH S =0 Here the more extra neutrons of the above unstable nuclides than those of Cd-116 make single bonds which lead to the decay.' ' ' ' STRUCTURE OF Cd-111, Cd-113, Cd-115, AND Cd-117 WITH S = +1/2 ' For understanding the structure of the above nuclides you must read my STRUCTURE OF Cd-111 . In this case using the diagram of Cd-96 with S =0 we see that the structure of them is given by adding extra neutrons with a total spin S =+1/2. For example the structure of the stable Cd-111 with S = +1/2 ,which has 15 extra neutrons, is giving by adding 8 extra neutrons of positive spins and 7 extra neutrons of negative spins giving a total S = +1/2. ' ''' '''SRUCTURE OF Cd-95 Cd-97, Cd-99, Cd-101, Cd-103, Cd-105, Cd-107, AND Cd-109 After a careful analysis I found that the structure of the above unstable nuclides is based on another structure of Cd-96 with S =+4. In this case the p37n37 and p38n38 of the diagram of Cd-96 with S =0 change their spins from S = -2 to S = +2 giving S = +4 . Particularly they move from -HSQ to +HSQ in order to make horizontal bonds with p38n38 and n40p40. Under this new structure of Cd-96, in the absence of a neutron with negative spin one gets the structure of Cd-95 with S = +9/2. That is S = +4 - 1(-1/2) = +9/2 On the other hand in the presence of the extra n49(+1/2) one gets the structure of the Cd-97 with S = +9/2. That is S = +4 + 1(+1/2) = +9/2. Whereas in the Cd-99 with S = +5/2 one adds 3 extra neutrons of negative spins. That is S = +4 + 3(-1/2) = +5/2 STRUCTURE OF Cd-119, Cd-121, Cd-123, Cd-125, Cd-127, AND Cd-129 WITH S =+3/2 After a careful analysis I found that the structure of the above unstable nuclides is based on another structure of Cd-96 having S =+2. In this case the p39n39 in the diagram of Cd-96 with S =0 changes the spin from S =-1 to S =+1 giving S = +2 . Particularly it moves from the -HSQ to +HSQ in order to make horizontal bonds with p38n38. Then in the presence of extra neutrons having a total S = -1/2 one gets the structure of the above nuclides. For example the Cd-119 with S = +3/2 (which has 13 extra neutrons) It has 6 extra neutrons of positive spins and 7 extra neutrons of negative spins giving a total S = -1/2 . That is S = +2 -1/2 =+3/2 Category:Fundamental physics concepts